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A particle of mass m is projected with a speed U at an angle $\theta$ with the horizontal. Find the torque of the weight of the particle about the point of projection when the particle is at the highest point .

\[\begin {array} {1 1} (a)\;z= \frac{w \times u^2 \sin 2 \theta}{g} & \quad (b)\;z= W \times \frac{u^2 \sin 2 \theta}{2g} \\ (c)\;z= \frac{w \times u^2 \sin ^2 \theta}{g} & \quad  (d)\;z= \frac{w \times u^2 \sin ^2 \theta}{2g} \end {array}\]

1 Answer

Torque= Force $\times $ perpendicular distance between lime of actions of force and the point of projection.
$z= w \times \large\frac{R}{2}$
answered Nov 25, 2013 by meena.p
 

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